2n^2+n=100

Solution for 2n^2+n=100 equation:

2n^2+n=100

We move all terms to the left:

2n^2+n-(100)=0

a = 2; b = 1; c = -100;
Δ = b2-4ac
Δ = 12-4·2·(-100)
Δ = 801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{801}=\sqrt{9*89}=\sqrt{9}*\sqrt{89}=3\sqrt{89}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{89}}{2*2}=\frac{-1-3\sqrt{89}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{89}}{2*2}=\frac{-1+3\sqrt{89}}{4}
$